A radioactive material decays by simultaneous emission of two particles with half-lives of 1400 yr and 700 yr, respectively. What will be the time after the which one-third of the material remains ? [Take, In 3 = 1.1]

The given situation can be shown as

Here, radioactive material X is decayed into two particles Y and Z with their respective decay constant, ${\mathrm{\lambda }}_{\mathrm{a}}\text{ and }{\mathrm{\lambda }}_{\mathrm{b}}$. It means that

$\because \mathrm{\lambda }=\frac{\ln 2}{{\mathrm{t}}_{1/2}}$

where, ${\mathrm{t}}_{1/2}\left(\mathrm{a}\right)=700\mathrm{yr}$

and ${\mathrm{t}}_{1/2}\left(\mathrm{b}\right)=1400\mathrm{yr}$

${\mathrm{\lambda }}_{\mathrm{a}}=\frac{\ln 2}{700}{\mathrm{yr}}^{-1}\text{ and }{\mathrm{\lambda }}_{\mathrm{b}}=\frac{\ln 2}{1400}{\mathrm{yr}}^{-1}$

$\begin{array}{c}\because {\mathrm{\lambda }}_{\text{total }}={\mathrm{\lambda }}_{\mathrm{a}}+{\mathrm{\lambda }}_{\mathrm{b}}\\ =\left(\frac{\ln 2}{700}+\frac{\ln 2}{1400}\right){\mathrm{yr}}^{-1}=\ln 2\left(\frac{1}{700}+\frac{1}{1400}\right){\mathrm{yr}}^{-1}\\ =\left(\frac{3\ln 2}{1400}\right){\mathrm{yr}}^{-1}\end{array}$

Suppose the initial number of radioactive nuclei was N₀.

$\therefore \mathrm{N}={\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda t}}$

where, N = number of nuclei present at time = t

and N₀ = number of nuclei present at time = 0

$\Rightarrow \frac{{\mathrm{N}}_0}{3}={\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda t}}\text{ or }\frac{{\mathrm{N}}_0}{3}={\mathrm{N}}_0{\mathrm{e}}^{-{\mathrm{\lambda }}_{\text{total }}\mathrm{t}}\Rightarrow \frac{1}{3}={\mathrm{e}}^{-{\mathrm{\lambda }}_{\text{total }}\mathrm{t}}$

Taking log on both the sides of above equation, we get

$\begin{array}{l}\ln \left(\frac{1}{3}\right)=\ln \left({\mathrm{e}}^{-{\mathrm{\lambda }}_{\text{total }}\mathrm{t}}\right)\\ \Rightarrow \ln \left(\frac{1}{3}\right)=-{\mathrm{\lambda }}_{\text{total }}\mathrm{t}\Rightarrow 1.1=\frac{3\times 0.693}{1400}\times \mathrm{t}\\ \Rightarrow \mathrm{t}\approx 740\mathrm{yr}\end{array}$